I bought several cheap heatsinks for ram chips (in particular to play with them with several Raspberrys constantly running overclocked) but I noticed that the volume of the "copper" ones does not match the metal standard density. In particular:

With 748 mm^3 (780 mm^3 the pad minus some holes), each heatsink should weight 6,6772464 grs (if the copper is pure, heheh... it is a shame that my country has so much copper but I have to buy copper parts machined in china ) but they weight 4.97 grs.

In the other hand, the aluminium ones are perfect, dimensions/weight match perfectly. So, how can I estimate a coefficient? I know the raspberry does not require heat sinks, and it is cheaper to buy some expensive ones, etc, etc but it is so easy to learn stuff sometimes and I used a lot of times these thermal coefficients in the university lab without really knowing how to calculate it.

My idea is to track the temperature change in one side of the heatsink using a peltier cell as heater bellow:

...and track 3 things based on ambient temp/humidity the time to get the same temp up there, retention time and time to cool down.

But I am unsure if this approach will be appropriate, should I take some consideration about the surface of the heatsink? density?

I did a little of that in a prior life and certainly couldn't do it now, but IIRC there were formulas for various simple forms (fins, pins, etc.) which could be combined into more complicated forms. At least in some cases these were based on experience and heuristics. Best bet would be the manufacturer's datasheet, it should give thermal resistance in °C/W and if it's a halfway decent datasheet it should also have charts showing the effect of airflow, ambient temperature, etc. Lacking that, I might try to find a datasheet for a heatsink with similar dimensions, use that to get a ballpark, and apply a generous safety factor.

Regarding the density calculations, if it doesn't come close to copper, how close is it to aluminum? Al is a much more common material for heatsinks than Cu.

Lastly, I suppose if a person knew how to do it properly, a known quantity of heat (power) could be applied to the heatsink, the temperature rise measured, and the thermal resistance calculated.

Regarding the density calculations, if it doesn't come close to copper, how close is it to aluminum? Al is a much more common material for heatsinks than Cu.

I am not sure what cheap metal you can mix with copper, but if they used copper+aluminum (but aluminum melting point it is very low ~650C vs ~1080C, so I am not sure a mix will be homogeneous), then is 64% copper

Is it not enough to heat it up to a certain temperature and measure the cool down temperature versus time?from my head it would be some negative exponential function like

Tt = Te + c1 * exp(c2 * -t). // Tt is temp at time t; Te is T environment, the temp it cools down to, c1 and c2 are constants depending on material form etc.

and c1 was the HTC? (I do not recall anymore)

Excel could help you find the equation if you have the sample data.

I want to use a similar formula I found in wikipedia, but then I was wondering the many variables the material form will have, it the shape was just a box, then it is easy, dissipation surfaces are the layers, but having these "pegs" increase the surface. The aluminium ones have another design:

And the copper ones seems less friendly with "directed wind flow":

Anyway copper is way too soft to shape it in thin shapes. To simplify the equation I can just consider the heatsink as a rectangle, so what is left is to determine how to apply the heat, and where to measure. Because for sure the contact surface with the heating element changes a lot the transference. The ceramic peltier cells are perfectly flat btw.

Yes the thermal resistance of a heat sink is dominated by the surface area, the material it is made from has little overall effect so copper or aluminum it makes little difference. You need to calculate the surface area by measuring all those notches.

I am not sure what cheap metal you can mix with copper, but if they used copper+aluminum (but aluminum melting point it is very low ~650C vs ~1080C, so I am not sure a mix will be homogeneous), then is 64% copper

Is it not enough to heat it up to a certain temperature and measure the cool down temperature versus time?from my head it would be some negative exponential function like

Tt = Te + c1 * exp(c2 * -t). // Tt is temp at time t; Te is T environment, the temp it cools down to, c1 and c2 are constants depending on material form etc.

and c1 was the HTC? (I do not recall anymore)

Excel could help you find the equation if you have the sample data.

I want to use a similar formula I found in wikipedia, but then I was wondering the many variables the material form will have, it the shape was just a box, then it is easy, dissipation surfaces are the layers, but having these "pegs" increase the surface. The aluminium ones have another design:

Still if you measure the cooling down curve, the graph would be (reasonably) identical to a box with the same surface area (assuming it can radiate freely)

I went through reliving my college physics and thermodynamics classes when writing the H,E,A.T. article here: http://www.instructables.com/id/LASER-HEAT/. But when all is said and done, a little bench work can can put those nasty equations into the closet :-)

Use a power-resistor and lab supply to heat the resistor to the same temperature as the Pi CPU when operating... Use an IR thermometer to measure the temp. Now, put the resistor on the heat sink oriented in the position it will operate and measure the temperature using the same power supply V & I. Note the stabilized temperature. Remove thee heatsink, reduce the I &E until the same temperature is reached w/o the heatsink as was recorded with the heatsink.

Your power supply values, I * E, gives the power in Watts between the no heatsink and heatsinked scenarios. Of course, our recordings are only approximations since mounting hardware, thermal grease, etc. Will have effects.

Farnell lists about 1120 passive heatsinks (natural convection ones), where about 215 are for BGA packages. The typical thermal resistance of an aluminum similarly shaped I see there is 25degC/Watt. So why to waste time with science?

Farnell lists about 1120 passive heatsinks (natural convection ones), where about 215 are for BGA packages. The typical thermal resistance of an aluminum similarly shaped I see there is 25degC/Watt. So why to waste time with science?